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Three discs $x, y$ and $z$ having radii $2 \mathrm{~m}, 2 \mathrm{~m}$ and $6 \mathrm{~m}$ respectively are coated on outer surfaces. The wavelength corresponding to maximum are power radiated by them respectively then
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The correct answer is:
$P_y$ is maximum
As we know that from Wien's displacement law $T \propto \frac{1}{\lambda_{\max }}$
From Stefan's Law, the emitted power is proportional to temperature raised to power four and directly poroportional to the surface area of the blackbody:
$P \propto A T^4$
So, $P \propto \frac{A}{\lambda_{\max }^4}$ where $A=4 \pi r^2$ is the surface area of the body with spherical shape of radius $r$.
Therefore,
$P_x: P_y: P_z=\frac{2^2}{3^4}: \frac{2^2}{4^4}: \frac{6^2}{5^4}=\frac{4}{81}: \frac{4}{16}: \frac{36}{625}=\frac{32}{648}: \frac{156}{624}: \frac{36}{625}$
$\Rightarrow P_y>P_x>P_z$
From Stefan's Law, the emitted power is proportional to temperature raised to power four and directly poroportional to the surface area of the blackbody:
$P \propto A T^4$
So, $P \propto \frac{A}{\lambda_{\max }^4}$ where $A=4 \pi r^2$ is the surface area of the body with spherical shape of radius $r$.
Therefore,
$P_x: P_y: P_z=\frac{2^2}{3^4}: \frac{2^2}{4^4}: \frac{6^2}{5^4}=\frac{4}{81}: \frac{4}{16}: \frac{36}{625}=\frac{32}{648}: \frac{156}{624}: \frac{36}{625}$
$\Rightarrow P_y>P_x>P_z$
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